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Rate of escape of conditioned Brownian motion
Orphée Collin, Francis Comets
To cite this version:
Orphée Collin, Francis Comets. Rate of escape of conditioned Brownian motion. 2021. �hal-03146283v2�
Rate of escape of conditioned Brownian motion
Orph´
ee Collin
1, Francis Comets
2April 4, 2021
1 Ecole Normale Sup´´ erieure, DMA, 45 Rue d’Ulm, 75005 Paris, France e-mail: orphee.collin@ens.psl.eu
2 Universit´e de Paris and LPSM, Math´ematiques, case 7012, F–75205 Paris Cedex 13, France
e-mail: comets@lpsm.paris
Abstract
We study the norm of the two-dimensional Brownian motion conditioned to stay outside the unit disk at all times. By conditioning the process is changed from barely recurrent to slightly transient. We obtain sharp results on the rate of escape to infinity of the process of future minima. For this, we introduce a renewal structure attached to record times and values. Additional results are given for the long time behavior of the norm.
Keywords: Brownian motion, Bessel process, conditioning, transience, Wiener mous-tache, regeneration, upper-class and lower-class, random difference equation
AMS 2020 subject classifications: 60K35, 60J60, 60J65, 60G17
Contents
1 Introduction 2
2 Main results 4
2.1 Results for the future minimum . . . 5
2.2 Long time behavior of R(t) . . . 5
3 Regenerative structure 6 3.1 Renewal times . . . 6
3.2 Description of a cycle . . . 7
3.3 Tail estimates for T . . . 9
3.4 Tail estimate for U . . . 12
4 Proofs for section 2.2 13
6 Proof of Theorem 2.2 19 6.1 Proof of (10). . . 20 6.2 Proof of (11) . . . 23
1
Introduction
This paper is devoted to the planar Brownian motion conditioned to stay outside the unit ball B(0, 1) at all times. Besides its own appeal from its fundamental character, this process has attracted a keen interest as being the elementary brick of the two-dimensional Brownian random interlacement recently introduced in [9]. By rotational symmetry, the norm R of the conditioned Brownian motion follows a stochastic differential equation in [1, ∞),
dR(t) = 1 R(t) ln R(t)+ 1 2R(t) dt + dB(t) (1)
with B a standard Brownian motion in R, and we can – and we will – restrict the study of the conditioned process to that of R itself since the angle obeys a diffusion subordinated to it. The two-dimensional Brownian motion is critically recurrent, but conditioning it outside the unit ball turns it into (delicately) transient. A natural question is the rate at which R(t) tends to ∞ as t → ∞, this is the object of the present paper. A measure of the reluctance of R to tend to infinity is given by the future minima process
M (t) = inf{R(s); s ≥ t} (2) which is non-decreasing to ∞ a.s. The corresponding model in the discrete case, the two-dimensional simple random walk conditioned to avoid the origin at all times, has motivated many recent papers. Estimates on the future minimum distance to the origin have been obtained in [23], we will use them as benchmarks. It is also shown that two independent conditioned walkers meet infinitely often although they are transient. The range of the walk, i.e. the set of visited sites, is studied in [12]: if a finite A ⊂ Z2 \ {0} is ”big enough and
well distributed in space”, then the proportion of visited sites is approximately uniformly distributed on [0, 1]. In [21] the explicit formula for the Green function is obtained, and a survey is given in Chapter 4 of [22].
For dimensions d ≥ 3, the random interlacement model has been introduced in [28] to describe the local picture of the visited set by a random walk at large times on a large d-dimensional torus, and similarly in [29], the Brownian random interlacement to describe the Wiener sausage around the Brownian motion on a d-dimensional torus. For dimension d = 2, the random interlacement model is the local limit of the visited set by the random walk around a point which has not been visited so far [7], and analogously, the Brownian random interlacement is the local limit of the Wiener sausage on the two-dimensional torus around a point which is outside the sausage [9]. Formally, the two-dimensional Brownian random interlacement is defined as a Poisson process of bi-infinite paths, which are rescaled instances of the so-called ”Wiener moustache”. The Wiener moustache is obtained by gluing two instances (for positive and negative times, see Figure 1 in [9]) of planar Brownian motion conditioned to stay outside the unit ball, which are independent except that they share the same starting
point (see Lemma 3.9 in [9]). Hence, the process we consider in this paper is the building brick of Brownian random interlacement in the plane. We also recall that the complement of the sausage around the interlacement has an interesting phase transition, changing from a.s. unbounded to a.s. bounded as the Poisson intensity is increased, see Th. 2.13 in [9] and [8] for the discrete case.
With a slight abuse of terminology, we say f (t) ≤ g(t) i.o. (infinitely often) if the set {t ≥ 0 : f (t) ≤ g(t)} is unbounded, and f (t) ≤ g(t) ev. (eventually) if the set {t ≥ 0 : f (t) ≤ g(t)} is a neighborhood of ∞ in R+.
We now give a short overview of some of our results on the rate of escape of R to infinity. They are consequences of the results in section 2.1.
Theorem 1.1. For g : R+ → R+ non-increasing such that (ln t)g(ln ln t) is non-decreasing,
P M (t) ≤ e(ln t)g(ln ln t) i.o. = 0 1 according to R∞ g(u)du < ∞ = ∞ .
This result with an integral condition has a flavor of Kolmogorov’s test (see, e.g., sect. 4.12 in [15]).
Theorem 1.2. The limit
K∗ = lim sup
t→∞
M (t) √
t ln ln ln t is a.s. constant, and 0 < K∗ < ∞.
Though we do not know the actual value of K∗ we can see that both theorems are much finer than the corresponding Theorem 1.2 of [23]. These two theorems together yield a precise version of the observation from [21] that the pathwise divergence of R to infinity occurs in a highly irregular way. The future minima process has been considered earlier, e.g. [17] and [18] for Bessel processes and for random walks, and [20] for positive self-similar Markov processes. Let us recall the similar result for transient Bessel processes. Denote by BESd the d-dimensional Bessel process, i.e. the solution of
dX(t) = d − 1
2X(t)dt + dB(t) , (3) that is the norm of the standard Brownian motion in Rd when d integer : then, by Th. 4.1 in [17],
for d > 2, lim sup
t→∞
min{BESd(s); s ≥ t}
√
2t ln ln t = 1. (4) An important (and beautiful) finding of our work is a renewal structure in Section 3 which allows sharp estimates. To illustrate that, let’s mention that we will find a sequence of relevant random variables Sn> 0 solving a random difference equation
Sn = αnSn−1+ βn, n ≥ 1, (5)
where the sequence (αn, βn)nis i.i.d. with positive coefficients, αn< 1 and βnwith logarithmic
usually addressed with exponential or power-law tail for βn [5], the case of logarithmic tail
has been also considered, see [16], [32], [3], and also [1] for a recent account. Interestingly, our model is critical in the perspective of the Markov chain Sn, in the sense that the actual value
of the constant c is precisely the transition from recurrence to transience for the chain. The paper is organized as follows. We give the main results in the next section. The regeneration structure is defined in Section 3 , together with the basic estimates, and ending with Remark 3.8 on the above random difference equation. In the next section we prove some results showing that R somewhat behaves at large times like the two-dimensional Bessel process. In Sections 5 and 6 we prove the two above theorems.
2
Main results
We first collect a few properties of the involved processes.
We start with some notations. Consider W a two-dimensional standard Brownian motion and denote by Px the law of W starting at x, cW a Brownian motion conditioned to stay
outside the unit ball, and denote by bPx its law starting at x, and R = |cW | its Euclidean norm
with Pr the corresponding law (r = |x|). In this paper we are mainly interested in P = P1.
The construction of the process starting from R(0) > 1 is standard from taboo process theory, and the one starting from R(0) = 1 is given in definition 2.2 of [9].
For a closed subset B of the state space of a process Y , we denote the entrance time τ (Y ; B) = inf{t ≥ 0 : Y (t) ∈ B}, and write for short τ (Y ; r) = τ (Y ; ∂B(0, r)) and also τ (r) = τ (R; r) when Y = R. The function h(x) = ln |x| is harmonic in R2 \ {0}, positive on R2 \ B(0, 1) and vanishes on the unit circle. Then, the law bP
x of the planar Brownian
motion W conditionned outside B(0, 1) is given by Doob’s h-transform of Px. By definition,
for A ⊂ C(R+, R) which is Fτ (r1)-measurable (1 < |x| = r < r1)
Pr(R ∈ A) = Px(|W | ∈ A τ (W, r1) < τ (W, 1)) (6) = Px(|W | ∈ A, τ (W, r1) < τ (W, 1)) × ln r1 ln |x| recalling that Px(τ (W, r1) < τ (W, 1)) = ln |x| ln r1 since ln |x| is harmonic in R 2\{0}.
Another remarkable property is Remark 3.8 in [9] : For all x /∈ B(1), ρ > 0, we have
b
Pxτ (cW ; B(y, ρ)) < ∞ →
1
2 as |y| → ∞ . The scale function for the process R is S(r) = ln r−1. Then, for 1 < a < r < b,
Pr[τ (b) < τ (a)] =
ln(r/a) × ln b
ln(b/a) × ln r. (7)
We refer to section 2.1 in [9] for more details on the many interesting properties of cW and R.
2.1
Results for the future minimum
With L(t) = ln(t ∨ 1) and ln(·) the natural logarithm, define ln1(t) = L(t), and for k ≥
2, lnk(t) = L(lnk−1(t)) so that lnk(t) = (ln ◦ . . . ◦ ln)(t) for t large.
Theorem 2.1. For g : R+→ R+ non-increasing such that (ln t)g(ln2t) is non-decreasing, we
have:
Z ∞
g(u)du < ∞ =⇒ a.s., M (t) ≥ e(ln t)g(ln2t) eventually, (8)
and
Z ∞
g(u)du = ∞ =⇒ a.s., M (t) ≤ e(ln t)g(ln2t) infinitely often. (9)
(Note that the second assumption is quite natural in view of the monotonicity of Mt.)
Theorem 1.1 is a direct consequence of the above theorem. This result with an integral condition is reminiscent of Kolmogorov’s test (see, e.g., sect. 4.12 in [15]), but the process M here is not Markov.
These estimates are stronger than the corresponding ones in Th. 1.2 of [23]. So are the following ones:
Theorem 2.2. There exist 0 < K0 < K < ∞ such that, almost surely,
M (t) ≤ Kpt ln3t eventually, (10)
and
M (t) ≥ K0pt ln3t infinitely often. (11)
Theorem 1.2 is essentially a reformulation of Theorem 2.2, it will be proved below Remark 6.2.
We recall the similar result (4) for transient Bessel processes: a.s. for all a <√2 < b, the future minima process min{BESd(s); s ≥ t} is eventually smaller than b√t ln
2t and infinitely
often larger than a√t ln2t.
Finally we mention that, for d > 2, min{BESd(s); s ≥ t} ≤ ε√2t ln2t i.o., a.s. for all ε > 0.
(See [17], P.349).
2.2
Long time behavior of R(t)
At large times the process R behaves like BES2. We emphasize that this is for the marginal
law, but not for the future minimum. We formulate here precise statements of these facts. It is well known that the random variable t−1/2BES2(t) converges to the Rayleigh distri-bution
dν(x) = xe−x2/21(0,∞)(x)dx (12)
as t → ∞. Similarly for R, we have Theorem 2.3. Let Z ∼ ν. As t → ∞, R(t) √ t law −→ Z .
Theorem 2.4 (Pointwise ergodic theorem). For all bounded continuous function f on (0, ∞), as t → ∞, 1 t Z et−1 0 f R(u) √ 1 + u 1 1 + udu −→ Z R f dν a.s. (13)
3
Regenerative structure
We fix a parameter r > 1. We construct a regenerative structure associated with the process R starting from R(0) = 1.
3.1
Renewal times
We define a random sequence (Hn, An, Tn)n≥0 by H0, T0 = 0, A0 = 1, then
H1 = inf{t > T0 : R(t) = r} A1 = inf{R(t); t ≥ H1} T1 = inf{t ≥ H1 : R(t) = A1} and for n ≥ 1, Hn+1 = inf{t > Tn : R(t) = rAn} An+1 = inf{R(t); t ≥ Hn+1} Tn+1 = inf{t ≥ Hn+1: R(t) = An+1} (14)
Since R is a continuous function with limt→∞R(t) = ∞ a.s., we see by induction that Tn< ∞
a.s, with Tn < Tn+1 and limn→∞Tn = ∞ a.s. The Tn are not stopping times, but they are
called renewal times for the following reasons.
Proposition 3.1. Let G1 = σ T1, (R(t)1{t < T1}; t ≥ 0). Then,
R(T1+ A21t)
A1
; t ≥ 0
has same law as R and is independent of G1 .
This proposition is the building brick of the Theorem 3.2. [Renewal structure] The sequence
R(Tn+ A2nt) An ; t ∈ 0,Tn+1− Tn A2 n n≥0
is independent and identically distributed with the law of (R(t); t ∈ [0, T1]).
In particular, since R(Tn+1) = An+1, the sequence
Tn+1− Tn A2 n ,An+1 An n≥0
is i.i.d. and distributed as (T1, A1). Therefore (Tn, An) can be written using i.i.d.r.v.’s, which
Proof. Proposition 3.1. Recall that Pr denotes the law of the process R with R(0) = r.
Observe that H1 is a stopping time, and denote by FH1 the sigma-field of events that occur
before time H1. By the strong Markov property,
under P1, (R(t + H1))t≥0is independent of FH1 and has the law Pr.
Moreover, by Theorem 2.4 in [31] (see also the proof of Lemma 3.9 in [9]), conditionnally on T1, (R(t); t ∈ [H1, T1]) and A1 = a, (R(T1+ t); t ≥ 0) has the same law as R starting from a
and conditionned to R(t) ≥ a, ∀t ≥ 0. By Remark 2.5 in [9], the latter law is equal to that of aR(·/a2) under P
1. Since G1 = σ(FH1; (R(t); t ∈ [H1, T1])) up to nul events, we obtain the
desired statement.
Proof. Theorem 3.2. By induction, Proposition 3.1 implies that for all n, the processR(Tn+A2nt) An ; t ≥ 0
is independent of Gn= σ Tn, (R(t); t < Tn) with the law of R. Then, the claim follows.
As a direct consequence we have discovered a simple representation of crucial times and points of the process.
Corollary 3.3. Define A0n+1= An+1 An , Tn+10 = Tn+1− Tn A2 n , n ≥ 0.
Then, (A0n, Tn0)n≥1 is an i.i.d. sequence with the same law as (A1, T1), and we have the
repre-sentation Tn = T10 + A 02 1T 0 2+ . . . + (A 0 1. . . A 0 n−1)2T 0 n An = A01. . . A 0 n n ≥ 1. (15)
3.2
Description of a cycle
Recall r > 1 is fixed. We will shorten the notations: (H, A, T ) = (H1, A1, T1). Recall that R
starts from R(0) = 1, hits r at H for the first time, and reaches its future minimum A ∈ (1, r) at time T . We also introduce its maximum B > r on [H, T ], as well as their logarithms U, V :
A = rU = min{R(t); t ≥ H}
B = rV = max{R(t); t ∈ [H, T ]}
see figure 1. It was shown in [9] that U is uniform on [0,1], but we can even compute the joint law of U, V . For 1 < a − h < a < r < b, we have by the strong Markov property
P(A ∈ [a − h, a], B > b) = Pr(τ (b) < τ (a)) × Pb min{R(t); t ≥ 0} ∈ [a − h, a] + o(h)
= ln(r/a) ln b ln(b/a) ln r × 1 a ln bh + o(h)
using (2.16) in [9] and that, for R started at b, min{R(t); t ≥ 0} has density (a ln b)−1 on (1, b). Hence (A, B) has a density given by the negative of the b-derivative of the dominant term as h & 0, i.e., pA,B(a, b) = 1 ab ln r ln(r/a) ln2(b/a) , 1 < a < r < b.
R(t) 0 1 H T t A r B
Figure 1: First cycle: A = rU, B = rV
By changing variables, it follows that (U, V ) has density
pU,V(u, v) =
1 − u
(v − u)21{0 < u < 1 < v} (16)
We recover that U is uniform on (0,1) and that V has density pV(v) = − ln 1 − 1/v − 1/v , v > 1.
It follows that for v ≥ 1,
P(V > v) = ∞ X n=1 1 n(n + 1)vn , (17) and then P(V > v) ∼ 1/(2v) as v → ∞.
We also need information on the cycle length T . For any s ≥ 1 we consider the hitting time by R starting at s of its absolute minimum, and denote by µs a r.v. with the same law:
µs∼ Ps arg min{R(t); t ≥ 0} ∈ ·
Recall that, under P, R(0) = 1.
Proposition 3.4. (i) We have
T = H + (T − H) , where H and (T − H) are independent with T − H law= µr.
(ii) For u ∈ (0, 1), the conditional law of T given U ≥ u is equal to the law of an independent sum H + r2uµ
Proof. (i) directly follows from the strong Markov property for the Markov process R and the stopping time H.
For (ii), we recall Remark 2.5 in [9]: for c > 1, denoting by Rc the diffusion R conditioned
to stay outside (1, c], we have
Rc(·) = cR(·/c2)
(Alternatively, this follows from R being the norm of conditioned Brownian motion (6) and from Brownian scaling.) Hence, for s ∈ R, again from the strong Markov property,
E1[eisT U ≥ u] = E1[eis(T −H+H) U ≥ u] = E1[eisH] × Er[eis(T −H) U ≥ u] = E1[eisH] × Er[eis×arg min{R(t);t≥0}
min{R(t); t ≥ 0} ≥ ru] = E1[eisH] × E[eisr
2uµ (r1−u)]
which proves the result.
3.3
Tail estimates for T
We need some estimates of the upper and lower tails of T , that we derive in this section. But first we state elementary comparisons of R and Bessel processes, see (3), that will be used all through the paper.
Proposition 3.5. (i) There exists a coupling of the processes R and BES2 starting at 1 such that
∀t ≥ 0, R(t) ≥ BES2(t) .
(ii) For δ > 0 there exists a coupling of the processes R and BES2+δ starting at 1 such that
for σ = sup{t ≥ 0; R(t) ≤ e2/δ},
∀s ≥ 0, R(σ + s) ≤ BES2+δ(σ + s) − BES2+δ(σ) + e2/δ .
Proof. It is well known [6] that the stochastic differential equation (3) has a strong solution, so we can couple the processes R and BES2, BES2+δ by driving equations (1) and (3) by the same Brownian motion B. Then, with x+ = max{x, 0} for x real, we have for all t > 0 and
all realization of B, d BES2(t) − R(t)+ = 1{BES2(t)≥R(t)} 1 2BES2(t) − 1 2R(t) − 1 R(t) ln R(t) dt ≤ 0
which implies (i) by integration. Similarly for (ii) we write the differential
d R(t) − BES2+δ(t)+ = 1{BES2+δ(t)≤R(t)} 1 2R(t) + 1 R(t) ln R(t)− 1 + δ 2BES2+δ(t) dt ≤ 0 for t ≥ σ.
We are now ready to start with the upper tail of T . Proposition 3.6. As t → ∞,
P(T ≥ t) ∼ ln r
ln t . (18)
More precisely, there exists constants t0 and C such that for all t ≥ t0,
1 −ln3t + C ln t ln r ln t ≤ P[T ≥ t] ≤ 1 + ln3t + C ln t ln r ln t. (19) Proof. We first obtain two preliminary estimates.
Upper bound: for 0 < ε < 1,
P(T ≥ t) = P T ≥ t, V ≥ ln t 2(1+ε) ln r +P T ≥ t, V < ln t 2(1+ε) ln r ≤ P V ≥ ln t 2(1 + ε) ln r + PR(s) ≤ t2(1+ε)1 , s ∈ [0, t] ≤ (1 + ε) ln r ln t + 1 5 2(1 + ε) ln r ln t 2 + C0exp −C1tε/(1+ε) (20)
for t ≥ t1 with t1 > 0 not depending on ε ∈ (0, 1). Indeed, to obtain the first term we have
used (17) in the form of P(V ≥ v) ≤ (1/2v) + (1/5v2) for large v. In order to obtain the second one, we first bound R(·) ≥ BES2(·), with BES2 started at 0 using Proposition 3.5, and
finally that there exist positive C0, C1 such that
∀t > 0, ∀ρ > 0, P BES2(s) ≤ ρ, s ∈ [0, t] ≤ C
0exp − C1
t
ρ2 , (21)
see e.g. exercise 1 p.106 in [27]. Lower bound: for 0 < ε < 1/2,
P(T ≥ t) ≥ P T − H ≥ t, V ≥ ln t 2(1 − ε) ln r = P V > ln t 2(1 − ε) ln r − P T − H ≤ t, V ≥ ln t 2(1 − ε) ln r ≥ P V > ln t 2(1 − ε) ln r − Pr τ (R, t2(1−ε)1 ) ≤ t ≥ (1 − ε) ln r ln t − C2exp −C3t ε/(1−ε) (22)
for t ≥ t2, with t2 > 0 not depending on ε ∈ (0,12). In (22) we have used (17) for the first
term, and we give details for the second one: for |x| = r > 1 by (6), we get for all t > 1,
Pr τ (R, t 1 2(1−ε)) ≤ t = P x τ (|W |, t2(1−ε)1 ) ≤ t τ (|W |, t 1 2(1−ε)) < τ (|W |, 1)
≤ Px τ (|W |, t2(1−ε)1 ) ≤ t × ln t 2(1 − ε) ln r ≤ C2exp −C3tε/(1−ε)
for some constant C2, C3 > 0 by the moderate deviation principle for Brownian motion.
For both the upper and lower bounds, we now choose
ε = εt=
ln3t + C4
ln t
with a constant C4. Provided the constant C4 is large enough, the terms C0exp −C1tεt/(1+εt)
and C2exp(−C3tεt/(1−εt)) are dominated by (ln t)−2. We then get (19) from (20) and (22),
taking any C > C4+ 4 ln r5 .
Finally, (18) is a direct consequence of (19). The proof is complete. We also need to control the lower tail of T .
Proposition 3.7. (i) For all ε ∈ (0, r − 1), there exists t0 > 0 such that for t ≤ t0,
P[T ≤ t] ≤ exp −(r − 1 − ε) 2 2t . (23)
(ii) For all ε > 0, there exists t1 > 0 such that for t ≤ t1, and all u ∈ [0, 1),
PT ≤ t|U ≥ u ≥ exp −(r − 1 + ε) 2 2t . (24)
Proof. (i) Setting a = 1 + ε/2 ∈ (1, r) and using the strong Markov property for the hitting time of a by R, we obtain P(T ≤ t) ≤ P1(τ (r) − τ (a) ≤ t) = Pa(τ (r) ≤ t) (6) = P(a,0)(τ (|W |, r) ≤ t τ (|W |; r) < τ (|W |, 1)) ≤ P(a,0)(τ (|W |, r) ≤ t) × ln r ln a .
Recalling large deviation results for Brownian motion in small time, e.g. section 6.8 of Ch. 5 in [2],
lim
t→0t ln P(a,0)(τ (|W |, r) ≤ t) = −
(r − a)2
2 , (25)
we see that the above upper bound implies (i).
(ii) Let t ≤ 1. By Proposition 3.4-(ii), and by comparing R and BES2 from Proposition
3.5 (i), we obtain P(T ≤ t|U ≥ u) ≥ P(H ≤ t − t2) × P(r2uµ(r1−u) ≤ t2) ≥ P(BES2(t − t2 ) ≥ r) × P µ(r1−u) ≤ t2 r2u
= P(1,0) |W (t − t2)| ≥ r ×
Pr1−u
arg min{R(s); s ≥ 0} ≤ θ, (26)
with θ = rt2u2 . We estimate the first term using again large deviation for Brownian motion in
small time [2]: for |x| < r,
lim
t→0t ln Px(|W |(t) ≥ r) = −
(r − |x|)2
2 . (27)
To estimate the second term in (26), note that R(θ) ≥ r1−u+√θ and R(s) ≥ r1−ufor all s ≥ θ implies that, Pr1−u-a.s., R achieves its minimum before time θ. Hence, by Markov property
and (7),
Pr1−u
arg min{R(s); s ≥ 0} ≤ θ ≥ Pr1−u
R(θ) ≥ r1−u+√θ× 1 − ln r 1−u ln(r1−u+√θ) ≥ PB(θ) ≥√θ× 1 − ln r 1−u ln(r1−u+√θ) ≥ P (B(1) ≥ 1)× t 2r ln r for large t,
arguing on the second line that R dominates Brownian motion by comparing the drift. Com-bined with (26) and (27), this completes the proof of (ii).
3.4
Tail estimate for U
Recall Hoeffding’s inequality [14], or Th. 2.8 in [4]: for b < 1, c > 1 and i ≥ 1,
P [2(U1+. . .+Ui) ≥ c.i] ≤ exp −
i
2(c − 1)
2, (28)
and
P [2(U1+. . .+Ui) ≤ b.i] ≤ exp −
i
2(1 − b)
2.
(29)
Remark 3.8 (The random difference equation (5)). Introduce the sequence
Sn =
Tn
A2 n
which is key in Section 6. In view of (15), we see that it solves the recursion Sn+1 = αn+1Sn+ βn+1
(i.e., (5) above), with
αn= (A0n) −2 , βn= Tn0 (A0 n)2 .
The bi-dimensional sequence (αn, βn), n ≥ 1, is i.i.d., and falls into the usual setup of random
difference equation. In our case, the following quantities exist a := E[ln α1] , b := lim
t→∞P[β1 > t] × ln t ,
and satisfy a < 0 (contractive case), 0 < b < ∞ (very heavy tail). Following [1], this prevents the Markov chain Snto be positive recurrent: though the contraction brings stability to the
pro-cess, yet occasional large values of βn overcompensate this behavior so that positive recurrence
fails to hold. In our case, we easily check from (18) that b = −a (= ln r)
in which case the Markov chain Sn is null recurrent, but in a critical manner: the chain is
transient if b > −a and null recurrent if b ≤ −a.
4
Proofs for section 2.2
We consider the process R from (1) on a geometric scale,
X(t) = e−t/2R(et−1) (30) and we observe that
β(t) = Z et−1 0 1 √ 1 + s dB(s)
is a standard Brownian motion by Paul L´evy’s characterization. We claim that X solves the SDE dX(t) = 1 2X(t) − X(t) 2 + 1 X(t) ln[et/2X(t)] dt + dβ(t) X(0) = R(0) . (31) Indeed, X(t) = e−t/2X(0) + e−t/2 Z et−1 0 1 2R + 1 R ln R(s) ds + e −t/2 B(et−1) = J (t) + K(t) + L(t) , with dJ (t) = −12J (t)dt, and dK(t) dt = − 1 2K(t) + 1 2X(t)+ 1 X(t) ln[et/2X(t)] , dL(t) = −1 2L(t)dt + e −t/2 dB(et−1) .
Moreover, we easily check the equality Z t 0 e−s/2dB(es−1) = Z et−1 0 1 √ 1 + u dB(u)
in the Gaussian space generated by B. Adding up terms, we see that X solves the SDE (31). Denote by bt, resp. b∞ the drift coefficient and its limit, given for x ∈ (0, ∞) by
bt(x) = 1 2x− x 2 + 1 x(ln x + t/2), b∞(x) = 1 2x − x 2, and by X(∞) the homogeneous diffusion
dX(∞)(t) = 1 2X(∞)(t) − X(∞)(t) 2 dt + dβ(t) . (32)
Following the approach of Takeyama [30], we state the following
Lemma 4.1. The diffusion X(t) = e−t/2R(et−1) is asymptotically homogeneous with homo-geneous limit X(∞), i.e, for all continuous f with compact support in (0, ∞) and all t > 0,
Ef (X(t + s))|X(s) = x −→ Exf (X(∞)(t))
as s → ∞ uniformly on compact subsets of (0, ∞).
Proof. It is easier to consider bX(t) = X(t) − e−t/2 which takes values in the fixed interval (0, ∞), and dX(s)(t) = bX(s + t). Then, the coefficients of the diffusion dX(s) converge to those
of X(∞), uniformly on compact subsets of (0, ∞), and the corresponding martingale problems
have a unique solution. Thus, Theorem 11.1.4 in [26] yields the desired result.
The process X(∞) is the transform X(∞)(t) = X(∞,2)(t) = e−t/2BES2(et − 1) of BES2 by
the rescaling and deterministic time-change (30). It is recurrent and ergodic on (0, ∞) with the Rayleigh law as invariant probability measure,
dν(x) = xe−x2/21(0,∞)(x)dx
A first consequence is that R marginally behaves like BES2. Corollary 4.2 (Convergence in law). Let Z ∼ ν. As t → ∞,
R(t) √
t
law
−→ Z .
Proof. Denote by Ps,t, Ps,t(∞)(0 ≤ s ≤ t) the Markov semi-groups associated to X and X(∞),
(Ps,tf )(x) = Ef (X(t))|X(s) = x , (P (∞)
s,t f )(x) = Ef (X(∞)(t))|X(∞)(s) = x ,
so that Ps,t(∞) = P0,t−s(∞) . For a bounded continuous f : (0, ∞) → R we write
P0,t+sf (x)− Z f dν = P0,s(Ps,s+tf )(x) − Z f dν =P0,s Ps,s+tf − P (∞) s,s+tf (x) + P0,s Ps,s+t(∞)f − Z f dν (x) ,
where both terms vanish as s, t → ∞, which is our claim. Indeed, by convergence of X(∞) to
equilibrium, Ps,s+t(∞)f −R f dν = P0,t(∞)f −R f dν → 0 uniformly on compact subsets of (0, ∞) as t → ∞ and Lemma 4.1 implies that Ps,s+tf − P
(∞)
s,s+tf → 0 uniformly on compact as s → ∞:
thus, we only need to prove tightness, i.e. that for all positive x,
infP0,s(1[ε,1/ε])(x); s ≥ 1 → 1 as ε → 0 . (33)
But this follows from the next two bounds
• R ≥ BES2 (see Proposition 3.5 (i)) which implies that X ≥ X(∞) ,
• sups≥1EX(s)2 ≤ sup
s≥1s−1ER(s)2 < ∞ that we explain now.
First recall from [9] that ln R1 is a martingale, and so, for all r > 1,
Er 1 ln R(t) = 1 ln r . (34)
By Itˆo’s formula,
d(R2) = 2 1 + 1 ln R(t) dt + 2R(t)dB(t) . (35) Thus, for all r > 1,
Er[R(t)2] = r2+ 2t 1 + 1 ln r .
We now consider the process starting from R(0) = 1. Integrating (35), we get
E1 R(t)2− r21 τ (r)<t = 2E1 Z t 0 1τ (r)<s 1 + 1 ln R(s)ds + Z t 0 1τ (r)<sR(s)dB(s) Markov = 2 Z t 0 E1 " 1τ (r)<sEr 1 + 1 ln R(·) ·=s−τ (r) # ds + 0 = 21 + 1 ln r E1 t − τ (r)+
by (34). Finally we obtain that
ER(t)2 ≤ r2+ 2t
1 + 1 ln r
for all r > 1. The corollary is proved.
Corollary 4.3 (Pointwise ergodic theorem). For all bounded continuous f on (0, ∞), as t → ∞, 1 t Z t 0 f (X(s))ds −→ Z R f dν a.s., or, equivalently, 1 t Z et−1 0 f R(u) √ 1 + u 1 1 + udu −→ Z R f dν a.s.
Proof. It is easy to check that, w.l.o.g., we can assume that f : (0, ∞) → R is non-decreasing. By the comparison principles of Proposition 3.5, we can couple the processes R, BES2, BES2+δ (δ > 0) starting at 1 such, a.s., for all t ≥ σ = sup{s > 0 : R(s) ≤ e2/δ} < ∞, we have
X(∞,2)(t) ≤ X(t) ≤ X(∞,2+δ)(t) −BES
2+δ(σ) − e2/δ
√
t .
By the pointwise ergodic theorem for X(∞,2) and X(∞,2+δ) and monotonicity of f , we derive
Z f dν ≤ lim inf t→∞ 1 t Z t 0 f (X(s))ds ≤ lim sup t→∞ 1 t Z t 0 f (X(s))ds ≤ Z f dνδ , where dνδ(x) = cδx1+δ/2e−x 2/2
1(0,∞)(x)dx is the invariant law of X(∞,2+δ). As δ vanishes, the
two extreme members coincide, ending the proof of the first statement. The second one follows by changing variables.
5
Proof of Theorem 2.1
Recall the representation (15) from Corollary 3.3, Tk= T10+ A 02 1T 0 2+ . . . + (A 0 1. . . A 0 k−1)2T 0 k , Ak= A01. . . A 0 k
with (Tk0, A0k)k≥1 an i.i.d. sequence with the same law as (T1, A1).
Fix r± with 1 < r− < r < r+< ∞. By Cram´er’s theorem [10], with probability one there
exists some finite random k0 such that for all k ≥ k0
r−k/2≤ A01. . . A 0
k = rU1+...+Uk ≤ r k/2 + .
In what follows we will use the rough bounds max i=1,...,kT 0 i ≤ Tk ≤ Tk0+ (k − k0) max i=1,...,kr i−1 + T 0 i . (36)
Lemma 5.1. There exists a constant c such that for all sequence (δ(k))k tending to 0, we have
P h k max i=1,...,kr i−1 + T 0 i ≥ e k/δ(k)i≤ cδ(k) eventually.
Proof. Fix a with 1 < a < e. Letting vk = a k δ(k) and t k = kr+kvk, we note that e k δ(k) ≥ t k
eventually since δ vanishes, and we have by independence
P[k max i=1,...,kr i−1 + T 0 i < tk] = Πi=1,...,kP[Ti0 < r+k−i+1vk]
From Proposition 3.6 there exists c1 > 0 such that for all t > 1
P(T1 ≥ t) ≤
c1
and since vk→ ∞ as k → ∞, we have for all large enough k, P[k max i=1,...,kr i−1 + T 0 i < tk] ≥ Πki=1 1 − c1 ln(r+k−i+1vk) = Πki=1 1 − c1 ln(ri +vk) ≥ exp −2c1 k X i=1 1 i ln r++ ln vk ! ≥ exp − 2c1 ln r+ ln k ln r++ ln vk ln vk = exp − 2c1 ln r+ ln(1 + ln r+ ln a δ(k)) ≥ 1 − cδ(k)
with c = 2c1/ ln a for all large k, since δ vanishes at ∞. This ends the proof.
Proof. Theorem 2.1, claim (8). Let
δ(t) = g(ln t), κ(i) = 2i, i ≥ 1, K = {κ(i) : i ≥ 1}. Define, for x ≥ 2, bxcK = max{k ∈ K : k ≤ x} = 2b(ln x)/(ln 2)c. Note that
x ≥ bxcK≥ x/2 . (37) First, since g is non-increasing,
X k∈K δ(k) = X i≥1 δ(k(i)) =X i≥1 g(ln k(i)) =X i≥1 g(i ln 2) ≤ 1 ln 2 X i≥1 Z i ln 2 (i−1) ln 2 g(t)dt = 1 ln 2 Z ∞ 0 g(t)dt < ∞ (38)
Fix a constant c2 > 0 to be chosen later and c3 = c−12 . Combining Borel-Cantelli’s lemma and
Lemma 5.1, we have a.s. k max i=1,...,kr i−1 + T 0 i < e
c2k/δ(k) for all k ∈ K large enough,
and, in addition to (36), we have for large k ∈ K,
Tk ≤ Tk0 +
k − k0
k e
since g is non-increasing. By integrability, g is vanishing at infinity, so the function f (t) = c3(ln t) g(ln2t)
is such that f (t) ≤ ln t eventually, and also g(ln2t) ≤ g(ln f (t)) by monotonicity. Thus, for
large k and t’s, k ≤ c3(ln t)δ(ln t) = f (t) implies that (40) k δ(k) = k g(ln k) ≤ f (t) g(ln f (t)) = c3(ln t)g(ln2t) g(ln f (t)) ≤ c3ln t.
Now, define random integers k(t) = max{k ∈ K; Tk ≤ t}, and note from (39) that a.s., for
large t we have k(t) ≥ max{k ∈ K; ec2δ(k)k ≤ t}. Then, a.s., for all large enough t,
Mt≥ MTk(t) = Ak(t) ≥ r k(t) 2 − ≥ r 1 2max{k∈K:e c2δ(k)k ≤t} − = r 1 2max{k∈K: k δ(k)≤c3ln t} − (using c3 = c−12 ) ≥ r 1 2max{k∈K:k≤f (t)} − (by (40)) = r 1 2bc3(ln t)δ(ln t)cK − ≥ rc34(ln t)δ(ln t) − (by (37))
Taking c3 = c−12 > 4/ ln r−, we conclude that a.s., M (t) ≥ e(ln t)g(ln2t) eventually, ending the
proof of (8).
We now turn to the proof of claim (9) of Theorem 2.1. We start with a lemma:
Lemma 5.2. Let (nk)k≥0 be a non-decreasing sequence of integers and (tk)k≥0 be a sequence
with tk > 1. Then,
X
k≥0
nk+1− nk
ln tk+1
= ∞ =⇒ a.s., Tnk ≥ tk infinitely often.
Proof. The events Ek = {maxi=nk+1,...,nk+1T 0
i ≥ tk+1}, k ≥ 0 are independent with Ek ⊂
{Tnk+1 ≥ tk+1}. Hence the conclusion holds as soon as these events occurs infinitely often
a.s. By the second Borel-Cantelli lemma, it suffices to show that the assumption implies P
k≥0P(Ek) = ∞. We use Proposition 3.6 and independence. The case when tk does not
tend to infinity is easily considered, so we assume from now on that k is large enough so that P(T ≥ tk+1) ≥ c/ ln tk+1 for some fixed constant c ∈ (0, ln r). Then, we can bound
P(Ek) = 1 − P(T ≤ tk+1)nk+1−nk ≥ 1 − 1 − c ln tk+1 nk+1−nk ≥ 1 − exp −c(nk+1− nk) ln tk+1
Proof. Theorem 2.1, claim (9). Let us consider
tk= ee k
, nk= bf (tk)c , f (t) = c3(ln t)g(ln2t)
with c3 > 0 to be fixed later. Note that f is non-decreasing by assumption. We have
X k≥0 nk+1− nk ln tk+1 =X k≥0 bf (tk+1)c − bf (tk)c ln(tk+1) =X k≥0 f (tk+1) − f (tk) ln(tk+1) + c4 = c3 X k≥0 g(k + 1) −1 eg(k) + c4
with a constant c4 which is finite since tkis increasing fast and the truncation error is bounded.
As in (38),P k≥0g(k) ≥ R∞ 0 g(t)dt = ∞, and n X k=0 g(k + 1) −1 eg(k) = g(n + 1) − 1 eg(0) + 1 −1 e n X k=1 g(k). ThereforeP k≥0 nk+1−nk
ln tk+1 = ∞. From Lemma 5.2 we obtain that a.s., Tnk ≥ tk i.o., which shows
that Mtk ≤ MTnk = Ank ≤ r nk + ≤ r f (tk) + .
Taking c3 < 1/ ln r+, we obtain the desired claim.
6
Proof of Theorem 2.2
We study the sequence
Sn= Tn A2 n = n X i=1 Ti0A2 i−1 A2 n = n X i=1 Ti0 r2(Ui+···+Un) ,
which can be written in the form
Sm = Sn r2(Un+1+···+Um) + S m n+1, (41) where, for 1 ≤ n < m, Sn+1m = m X i=n+1 Ti0 r2(Ui+···+Um).
The point is that, in (41), Sn and Sn+1m are independent, with Sn+1m equal to Sm−n in law.
We study the convergence/divergence of the series P
n≥1P[Sn≤ tn], with tn of the form
tn=
β ln2n
∧ 1 (42)
6.1
Proof of (10).
Let (i(n))i≥1 be a sequence of integers such that 1 ≤ i(n) ≤ n and (c(n)i )i=i(n)+1,...,n,n≥1 be a
doubly-indexed sequence of real parameters with c(n)i > 1, to be fixed later on.
Upper bound:
From (41) we have P[Sn ≤ tn] ≤ P T10 r2(U1+···+Un) ≤ tn, S n 2 ≤ tn ≤ P T10 r2(U1+···+Un) ≤ tn, S n 2 ≤ tn, 2(U1+ · · · + Un) ≤ c(n)n .n + P[2(U1+ · · · + Un) > c(n)n .n] ≤ P[T10 ≤ tnrc (n) n .n, Sn 2 ≤ tn] + P[2(U1+ · · · + Un) > c(n)n .n] ≤ P[T ≤ tnrc (n) n .n] × P[S n−1 ≤ tn] + P[2(U1+ · · · + Un) > c(n)n .n].Iterating the estimate,
P[Sn−1 ≤ tn] ≤ P[T ≤ tnrc (n) n−1.(n−1)] × P[S n−2≤ tn] + P[2(U1+ · · · + Un−1) > c (n) n−1.(n−1)],
and so on down to i(n)+ 1, we obtain
P[Sn ≤ tn] ≤ n Y i=i(n)+1 P[T ≤ tnrc (n) i .i] × P[Si(n) ≤ tn] + n X i=i(n)+1 n Y j=i+1 P[T ≤ tnrc (n) j .j] !
× P[2(U1+. . .+Ui) > c(n)i .i].
(43)
Choice of i
(n)and the c
(n)iLet i(n) = bln2nc and for i(n)+ 1 ≤ i ≤ n,
c(n)i = 1 + r
8
i(ln i + ln2n). (44) By (42), we have for i(n)+ 1 ≤ i ≤ n and large n,
ln P[T ≤ tnrc (n) i .i] ≤ ln P[T ≤ rc (n) i .i] ≤ −P[T ≥ rc(n)i .i] ≤ − 1 c(n)i .i + εn,i,1 (by (19))
≤ −1
i + εn,i,2 (by (44)), with error terms
εn,i,1 = ln2 c(n)i .i ln r+ C c(n)i .i2ln r , εn,i,2 = εn,i,1+ r 8 i3(ln i + ln2n) .
One can check that supnPn
i=i(n)+1εn,i,2 < ∞, so for some positive constant D, for n large
and i(n)≤ i ≤ n, n Y j=i+1 P[T ≤ tnrc (n) j .j] ≤ exp − n X j=i+1 1 j + n X j=i+1 εn,j,2 ! ≤ D exp− lnn i ≤ Di n. (45)
Combining this with (28), we get for n large and i(n)+ 1 ≤ i ≤ n, n Y j=i+1 P[T ≤ tnrc (n) j .j] ! × P[2(U1+. . .+Ui) > c (n) i .i] ≤ D i n exp(−4(ln i + ln2n)) = D i3n(ln n)4.
Thus, the series P an, with
an= n X i=i(n)+1 n Y j=i+1 P[T ≤ tnrc (n) j .j] !
× P[2(U1+. . .+Ui) > c(n)i .i],
is convergent.
Choice of tn
To conclude, we need to take care of the first term in the right-hand side of (43). Recall tn
from (42) (we will assume n large so that ln2n ≥ β), and fix an integer i1 ≥ 1. For 1 ≤ i ≤ i1,
applying (23) we get as n → ∞, for any ∈ (0, r − 1),
P[T ≤ tnr2i] ≤ exp −(r − 1 − ) 2 2βr2i ln2n ,
and then, for n large,
P[Si(n) ≤ tn] ≤ P[Ti0 ≤ tnr2i, i = 1, . . . , i1] = i1 Y i=1 P[T ≤ tnr2i]
≤ exp − i1 X i=1 (r − 1 − )2 2βr2i ln2n ! ≤ exp −(r − 1 − ) 2 2β 1 r2 1 − r12 i1 1 − r12 ln2n ! ≤ (ln n)− (r−1−)2 2β(r2−1) 1−(1 r2) i1 . Using (45) we will bound
n Y i=i(n)+1 P[T ≤ tnrc (n) i .i] × P[Si(n) ≤ tn] ≤ D i(n) n (ln n) −(r−1−)2 2β(r2−1)(1−( 1 r2) i1) , where i(n)= bln2nc. As soon as β < (r−1)
2(r+1), there exists some integer i1 and some ∈ (0, r − 1)
such that (r − 1 − )2 2β(r2− 1) 1 − 1 r2 i1! > 1,
and combining (43) with P
nan < ∞, we obtain P P(Sn ≤ tn) < ∞. i.e., X n≥1 P[Tn≤ A2ntn] < ∞. Conclusion
Let β < 2(r+1)(r−1). It follows from Borel-Cantelli’s lemma that a.s., eventually
Tn≥
βA2n ln2n
.
Now, for Tn≤ t ≤ Tn+1, if n is large enough,
Mt≤ MTn+1 = An+1≤ rAn≤ r p β−1T nln2n ≤ r p β−1t ln 2n,
and since we have Tn≥ βA 2 n ln2n ≥ r
n 2
− for n large enough, we have t ≥ r n 2 −, and n ≤ ln r2 ln t−. Finally, Mt≤ r s β−1t ln 2 2 ln t ln r− .
Hence, we have proved (10) with any K > r q
2(r+1)
(r−1). Optimizing on r > 1, we get the result
6.2
Proof of (11)
We start by proving that it suffices to show divergence of the series introduced above (42): Lemma 6.1. Let β0 = inf{β > 0 :
P
nP(Sn≤ lnβ2n) = ∞}. Then
lim inf
n Snln2n = β0 a.s.
Proof. For all β < β0, we have PnP(Sn ≤ lnβ
2n) < ∞ and the first Borel-Cantelli’s lemma
shows that lim infnSnln2n ≥ β0. To prove the reverse inequality we proceed by steps:
• First step: For any non-increasing sequence (tn)n,
X n≥1 P[Sn ≤ tn] = ∞ =⇒ P(Sn≤ tni.o.) ≥ 1 4. Indeed, for 1 ≤ n ≤ m, P[Sn ≤ tn, Sm ≤ tm] ≤ P[Sn≤ tn, Sn+1m ≤ tm] = P[Sn ≤ tn] × P[Sn+1m ≤ tm] = P[Sn ≤ tn] × P[Sm−n ≤ tm] ≤ P[Sn≤ tn] × P[Sm−n ≤ tm−n],
since tm ≤ tm−n. Now, for k ≥ 1,
X 1≤n<m≤k P[Sn≤ tn, Sm ≤ tm] ≤ X 1≤n<m≤k P[Sn≤ tn] × P[Sm−n≤ tm−n] ≤ X 1≤n,m≤k P[Sn≤ tn] × P[Sm ≤ tm].
For all k large enough we have Pk
n=1P[Sn ≤ tn] ≥ 2, and then for all 1 ≤ n ≤ k,
X 1≤m≤k,m6=n P[Sm ≤ tm] ≥ 2 − P[Sn≤ tn] ≥ P[Sn≤ tn]. Therefore, X 1≤n,m≤k P[Sn≤ tn] × P[Sm ≤ tm] ≤ 2 X 1≤n,m≤k,n6=m P[Sn≤ tn] × P[Sm ≤ tm] = 4 X 1≤n<m≤k P[Sn ≤ tn] × P[Sm ≤ tm]
Cochen-Stone’s theorem [19] – a variant of Borel-Cantelli’s lemma – yields
P[Sn≤ tni.o.] ≥ lim sup k≥1 P 1≤n<m≤kP[Sn≤ tn] × P[Sm≤ tm] P 1≤n<m≤kP[Sn ≤ tn, Sm ≤ tm] ≥ 1 4,
which concludes this step.
• Second step: Let’s introduce the σ-fields
Ak = σ((A0n, T 0 n); n ≥ k), k = 1, 2 . . . , T = \ k≥1 Ak.
By Kolmogorov 0–1 law and independence of the sequence ((A0n, Tn0); n ≥ 1), every element A of the tail field T has P(A) ∈ {0, 1}. Fix β ≥ 0 and introduce the events
E = {lim inf n Snln2n ≤ β}, Ek= {lim infn S n+k k+1 ln2n ≤ β}, and Ω0 = { lim n→∞ ln2n r2(U1+...+Un) = 0}.
Note that E = E0 and that P(Ω0) = 1. Since, by definition,
Sk+1n+k+1= T 0 k+1 r2(Uk+1+...+Un+k+1) + S n+k+1 k+2 ,
we see that the two sets Ek and Ek+1 coincide on Ω0, for all k ≥ 0. Denoting the common
intersection by
b
E = E\Ω0 = Ek
\ Ω0 ,
we see that bE belongs to T and then has probability equal to 0 or 1. The similar 0–1 law holds for E which is equal to bE up to a negligible set.
• Final step: For any β > β0, the series
P
nP(Sn≤ tn) with tn = β/ ln2n is diverging. By
the first step, the probability P[Sn≤ tni.o.] ≥ 1/4, and by the second one is equal to 1. Thus
lim infnSnln2n ≤ β a.s., for all such β’s. The lemma is proved.
Remark 6.2. We have followed the approach of the renewal structure to get the 0–1 law, with the advantage to keep the paper self-contained. A tempting alternative would be to show that the tail σ-field of R is trivial; we mention the illuminating survey [24] on the tail σ-field of a diffusion.
Anticipating on the proof of (11) we now give a short proof of Theorem 1.2.
Proof. It is not difficult to check the criteria of [11] or [25] for triviality of the tail σ-field of one-dimensional diffusion (see Theorem 3 in [24]). Then, K∗ = lim supt→∞√M (t)
t ln3t is a.s.
constant. Then, (10) and (11) show that K∗ is positive and finite. To continue the proof of (11) we need an intermediate result.
Lemma 6.3. For all α0 > 0, there exists β > 0 such that, for all n large enough,
PSbα0ln2nc ≤
β ln2n
≥ 1 ln n.
Proof. Clearly, it suffices to prove that for v > 0, there exists u > 0 such that, for all large n we have, P[Sn ≤ u n] ≥ 1 evn. (46)
Indeed, substituting v, n in (46) by α−10 , bα0ln2nc shows that any β > u/α0 fulfills the
state-ment of the lemma.
To show (46), we fix some b ∈ (0, 1) (b will be chosen small later on), and we note that: Ui ≥ b and Ti0 ≤ un(r
b− 1)rb(n−i+1) for all i = 1, . . . , n
imply that Sn= n X i=1 Ti0 r2(Ui+···+Un) ≤ n X i=1 u n(r b− 1)rb(n−i+1) r2b(n−i+1) ≤ u n. Then, P[Sn≤ u n] ≥ n Y i=1 P[Ui ≥ b, Ti0 ≤ u n(r b− 1)rb(n−i+1) ] = (1 − b)n n Y i=1 P[Ti0 ≤ u n(r b− 1)rb(n−i+1)|U i ≥ b] = (1 − b)n n Y i=1 P[T ≤ u n(r b − 1)rbi|U ≥ b] (47)
By Proposition 3.7, we can find t0 > 0 and ρ > 0 such that, for t ≤ t0,
P[T ≤ t|U ≥ b] ≥ exp(− ρ t).
Now, we fix some t1 > t0, we will bound the factors in (47) as follows:
For ln(t1 n u(rb−1)) b ln r ≤ i ≤ n : P[T ≤ u n(r b− 1)rbi |U ≥ b] ≥ P[T ≤ t1|U ≥ b], for ln(t0 n u(rb−1)) b ln r ≤ i ≤ ln(t1 n u(rb−1)) b ln r : P[T ≤ u n(r b− 1)rbi |U ≥ b] ≥ P[T ≤ t0|U ≥ b], and for 1 ≤ i ≤ ln(t0 n u(rb−1)) b ln r : P[T ≤ u n(r b− 1)rbi|U ≥ b] ≥ exp −ρ n u(rb− 1) 1 rbi .
With this choice, the estimate (47) becomes
P[Sn≤ u n] ≥ (1 − b) n × P[T ≤ t1|U ≥ b]n× P[T ≤ t0|U ≥ b] ln(t1 t0) b ln r +1
× bln(t0 n u(rb−1)) b ln r c Y i=1 exp −ρ n u(rb− 1) 1 rbi ≥ (1 − b)n × P[T ≤ t1|U ≥ b]n× P[T ≤ t0|U ≥ b] ln(t1 t0) b ln r +1 × exp −ρ n u(rb− 1)2
From this we derive the claim (46) by taking b small, u and t1 large. This ends the proof of
the lemma.
Proof. Theorem 2.2, claim (11). Similarly to the proof of (10), we let tn= lnβ2n∧1, (i(n))n≥1be
a sequence of integers, and (b(n)i )i=i(n)+1,...n,n≥1be a doubly-indexed sequence with 0 < b(n)i < 1,
given by b(n)i = 1 − r 8 i(ln i + ln2n) , for i (n)+ 1 ≤ i ≤ n, i(n)= bα 0ln2nc
with α0 large (take α0 > 8 so that b (n)
i > 0 for n large).
This time, we need an extra doubly-indexed, positive sequence (s(n)i )i=i(n)+1,...,n,n≥1 such
that for n large
n
X
i=i(n)+1
s(n)i ≤ tn.
(Note that this implies s(n)i ≤ 1.) Similarly, using (41) we estimate
P[Sn≤ tn] ≥ P T10 r2(U1+···+Un) ≤ s (n) n , S n 2 ≤ tn− s(n)n ≥ P T10 r2(U1+···+Un) ≤ s (n) n , S n 2 ≤ tn− s(n)n , 2(U1+ · · · + Un) ≥ b(n)n .n ≥ PhT10 ≤ s(n) n r b(n)n .n, Sn 2 ≤ tn− s(n)n , 2(U1+ · · · + Un) ≥ b(n)n .n i ≥ PhT10 ≤ s(n) n r b(n)n .n, Sn 2 ≤ tn− s(n)n i − P[2(U1+ · · · + Un) < b(n)n .n] ≥ PhT ≤ s(n)n rb(n)n .n i × PSn−1 ≤ tn− s(n)n − P[2(U1+ · · · + Un) < b(n)n .n].
We iterate the procedure,
P[Sn−1≤ tn− s(n)n ] ≥ P h T ≤ s(n)n−1.rb(n)n−1.(n−1) i × PhSn−2≤ tn− s(n)n − s (n) n−1 i − P[2(U1+ · · · + Un−1) < b(n)n−1.(n − 1)],
and so on down to i(n). We obtain
P[Sn≤ tn] ≥ n Y i=i(n)+1 P h T ≤ s(n)i rb(n)i .i i × P Si(n) ≤ tn− n X i=i(n)+1 s(n)i
− n X i=i(n)+1 n Y j=i+1 P h T ≤ s(n)j rb(n)j .j i ! × P[2(U1+. . .+Ui) < b (n) i .i]. (48)
Using s(n)i ≤ 1 and b(n)i < 1, we have, for n large and i(n)+ 1 ≤ i ≤ n : n Y j=i+1 P[T ≤ s(n)j r b(n)j .j] ≤ n Y j=i+1 P[T ≤ rj] ≤ exp − n X j=i+1 P[T ≥ rj] ! ≤ exp − n X j=i+1 1 j − ln2(j ln r) + C j2ln r ! (by (19)) ≤ D0i n, for some positive constant D0.
As we did for the series P
nan, cf. below (45) except for using (29) instead of (28), we
easily see that the series P
na 0 n, with a0n= n X i=i(n)+1 n Y j=i+1 P[T ≤ s(n)j r b(n)j .j ] ! × P[2(U1+. . .+Ui) < b (n) i .i],
is finite. Now, we choose
s(n)i = 1 i3 ,
and we start to bound from below the product
n Y i=i(n)+1 P[T ≤ s(n)i r b(n)i .i] = exp n X i=i(n)+1 ln(1 − P[T ≥ s(n)i r c(n)i i])
Observe that, by taking α0 > 16, we have b (n)
i ∈ (1/2, 1) for all large n and i ∈ [i
(n)+ 1, n],
and also that
inf{s(n)i rb(n)i .i; i(n)≤ i ≤ n} ≥ rα02 ln2n for large n, (49)
which tends to ∞ as n → ∞. For i(n)+ 1 ≤ i ≤ n and n large, in view of (49) we have (using
− ln(1 − u) ≤ u + u2 for small u > 0 and 1
1−u ≤ 1 + 2u for 0 < u < 1 2) − ln(1 − P[T ≥ s(n)i r b(n)i .i ]) ≤ P[T ≥ s(n)i r b(n)i .i] + ε0 n,i,1 ≤ ln r lns(n)i rb(n)i .i + ε 0 n,i,2 (by (19)) = 1 b(n)i .i +ln s (n) i ln r + ε0n,i,2
≤ 1 b(n)i .i + ε 0 n,i,3 ≤ 1 i + ε 0 n,i,4 ,
with error terms
ε0n,i,1 = PhT ≥ s(n)i rb(n)i .i i2 , ε0n,i,2 = ε0n,i,1 + 1 ln r × ln3 s(n)i rb(n)i .i+ C b(n)i .i +ln s (n) i ln r 2 , ε0n,i,3 = ε0n,i,2 − 2 ln s (n) i b(n)i .i2ln r , ε0n,i,4 = ε0n,i,3+ 2 r 8 i3(ln i + ln2n).
One can check that supnPn
i=i(n)+1ε0n,i,4 < ∞, so for some positive constant D00, for large n, n Y i=i(n)+1 P[T ≤ s(n)i r b(n)i .i] ≥ exp − n X i=i(n)+1 1 i + ε 0 n,i,4 ≥ D00i (n) n . (50)
Finally, consider the term
P Si(n) ≤ tn− n X i=i(n)+1 s(n)i .
Note that tn−Pni=i(n)+1s (n) i = β ln2n − Pn i=i(n)+1 i13 ≥ β ln2n − 1
2 i(n)2, which implies that for all
β0 < β, tn− Pn i=i(n)+1s (n) i ≥ β0
ln2n for large n, and then
P Si(n) ≤ tn− n X i=i(n)+1 s(n)i ≥ P Si(n) ≤ β0 ln2n .
Now, we are ready to conclude the proof: Fix α0 > 16, and let β0 be associated to α0 by
Lemma 6.3. Then, P Si(n) ≤ β0 ln2n ≥ 1 ln n, and for tn = (β/ ln2n) ∧ 1 with β > β0, using (50),
n Y i=i(n)+1 P[T ≤ s(n)i r b(n)i .i] × P Si(n) ≤ tn− n X i=i(n)+1 s(n)i ≥ D 00i(n) n × 1 ln n.
Using now (48) andP
na 0
n< ∞ we obtain
P
n≥1P[Sn≤ tn] = ∞. By Lemma 6.1 we have a.s.,
Tn ≤
βA2 n
ln2n
i.o.
i.e., An ≥ pβ−1Tnln2n. Since, for all large n, βA 2 n ln2n ≤ r n +, we see that Tn ≤ rn+, so n ≥ ln Tn ln r+, and also MTn = An ≥ s β−1T nln2 ln Tn ln r+ .
Finally, for some (small) K0 > 0, with probability one, Mt ≥ K0
√
t ln3t i.o. The proof of (11)
is complete.
Acknowledgements: FC is partially supported by ANR SWIWS.
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